Integrand size = 21, antiderivative size = 65 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {(5 a+4 b) \tan (e+f x)}{5 f}+\frac {b \sec ^4(e+f x) \tan (e+f x)}{5 f}+\frac {(5 a+4 b) \tan ^3(e+f x)}{15 f} \]
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Time = 0.05 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {4131, 3852} \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {(5 a+4 b) \tan ^3(e+f x)}{15 f}+\frac {(5 a+4 b) \tan (e+f x)}{5 f}+\frac {b \tan (e+f x) \sec ^4(e+f x)}{5 f} \]
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Rule 3852
Rule 4131
Rubi steps \begin{align*} \text {integral}& = \frac {b \sec ^4(e+f x) \tan (e+f x)}{5 f}+\frac {1}{5} (5 a+4 b) \int \sec ^4(e+f x) \, dx \\ & = \frac {b \sec ^4(e+f x) \tan (e+f x)}{5 f}-\frac {(5 a+4 b) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{5 f} \\ & = \frac {(5 a+4 b) \tan (e+f x)}{5 f}+\frac {b \sec ^4(e+f x) \tan (e+f x)}{5 f}+\frac {(5 a+4 b) \tan ^3(e+f x)}{15 f} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.94 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {a \left (\tan (e+f x)+\frac {1}{3} \tan ^3(e+f x)\right )}{f}+\frac {b \left (\tan (e+f x)+\frac {2}{3} \tan ^3(e+f x)+\frac {1}{5} \tan ^5(e+f x)\right )}{f} \]
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Time = 0.48 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89
method | result | size |
derivativedivides | \(\frac {-a \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )-b \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )}{f}\) | \(58\) |
default | \(\frac {-a \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )-b \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )}{f}\) | \(58\) |
parts | \(-\frac {a \left (-\frac {2}{3}-\frac {\sec \left (f x +e \right )^{2}}{3}\right ) \tan \left (f x +e \right )}{f}-\frac {b \left (-\frac {8}{15}-\frac {\sec \left (f x +e \right )^{4}}{5}-\frac {4 \sec \left (f x +e \right )^{2}}{15}\right ) \tan \left (f x +e \right )}{f}\) | \(60\) |
parallelrisch | \(\frac {\left (50 a +40 b \right ) \sin \left (3 f x +3 e \right )+\left (10 a +8 b \right ) \sin \left (5 f x +5 e \right )+40 \sin \left (f x +e \right ) \left (a +2 b \right )}{15 f \left (\cos \left (5 f x +5 e \right )+5 \cos \left (3 f x +3 e \right )+10 \cos \left (f x +e \right )\right )}\) | \(85\) |
risch | \(\frac {4 i \left (15 a \,{\mathrm e}^{6 i \left (f x +e \right )}+35 a \,{\mathrm e}^{4 i \left (f x +e \right )}+40 b \,{\mathrm e}^{4 i \left (f x +e \right )}+25 a \,{\mathrm e}^{2 i \left (f x +e \right )}+20 b \,{\mathrm e}^{2 i \left (f x +e \right )}+5 a +4 b \right )}{15 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}\) | \(87\) |
norman | \(\frac {-\frac {2 \left (a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {2 \left (a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{f}+\frac {8 \left (2 a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 f}+\frac {8 \left (2 a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{3 f}-\frac {4 \left (25 a +29 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{15 f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{5}}\) | \(119\) |
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Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.86 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {{\left (2 \, {\left (5 \, a + 4 \, b\right )} \cos \left (f x + e\right )^{4} + {\left (5 \, a + 4 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )^{5}} \]
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\[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \sec ^{4}{\left (e + f x \right )}\, dx \]
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Time = 0.19 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.66 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {3 \, b \tan \left (f x + e\right )^{5} + 5 \, {\left (a + 2 \, b\right )} \tan \left (f x + e\right )^{3} + 15 \, {\left (a + b\right )} \tan \left (f x + e\right )}{15 \, f} \]
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Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {3 \, b \tan \left (f x + e\right )^{5} + 5 \, a \tan \left (f x + e\right )^{3} + 10 \, b \tan \left (f x + e\right )^{3} + 15 \, a \tan \left (f x + e\right ) + 15 \, b \tan \left (f x + e\right )}{15 \, f} \]
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Time = 18.42 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.65 \[ \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^5}{5}+\left (\frac {a}{3}+\frac {2\,b}{3}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (a+b\right )\,\mathrm {tan}\left (e+f\,x\right )}{f} \]
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